Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A 22 kW motor running at 1450 rpm drives a pinion with 24 teeth and a module of 4 mm. Find the tangential force Ft on the pinion teeth.
P = 22 kWn = 1450 rpmN1 = 24m = 4 mm
Ft = N
2.Gears & gear trainsHard
An open belt connects an 80 mm driving pulley at 960 rpm to a 500 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 80 mmn1 = 960 rpmd2 = 500 mmC = 500 mmμ = 0.25T1 = 2000 N
P = kW
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 250 mm rotate 1.48° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 20 N·md = 15 mmL = 250 mmφ = 1.48°
G = GPa
4.Torsion of shaftsHard
A solid shaft must transmit 30 kW at 750 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 30 kWN = 750 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −10 MPa, σᵧ = 10 MPa and τₓᵧ = −20 MPa (tension positive). The element is rotated 30° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −10 MPaσy = 10 MPaτxy = −20 MPaθ = -30°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 50 MPa, σᵧ = 90 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 20° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 50 MPaσy = 90 MPaτxy = 40 MPaθ = 20°
τθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.
segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of aluminium (diameter 60 mm, length 400 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 400 mm, E = 70 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 30 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 30 kNn = 2L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 4 cycles later by a peak of 3.2 mm. Find the damping ratio ζ.
x0 = 40 mmx4 = 3.2 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 5 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 40 kgk1 = 12 kN/mk2 = 5 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 3018 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Torque on the pinion
ω = 2π·1450/60 = 151.8 rad/s
T = P/ω = 22000/151.8 = 144.9 N·m
Pitch diameter
d = m·N = 4·24 = 96 mm
Tangential force
Ft = 2T/d = 2·144.9/0.096 = 3018 N
Answer
Ft = 3018 N
2.Gears & gear trains
Answer: P = 3.488 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Angle of wrap
θ = 2.275 rad
Slack-side tension
eμθ = e0.5687 = 1.766, T₂ = 2000/1.766 = 1133 N
Power
P = (2000 − 1133)·4.021 = 3.488 kW
Answer
P = 3.488 kW
3.Torsion of shafts
Answer: G = 38.95 GPa
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Polar moment and angle
J = π·15⁴/32 = 4970 mm⁴
φ = 1.48·π/180 = 0.02583 rad
Shear modulus
T = 20 N·m = 20000 N·mm
G = TL/(Jφ) = 20000·250/(4970·0.02583) = 38946 MPa
G = 38.95 GPa
Answer
G = 38.95 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: d = 28.97 mm
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Torque
ω = 2πN/60 = 2π·750/60 = 78.54 rad/s
T = P/ω = 30000/78.54 = 382 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·381972/(π·80)) = 28.97 mm
Answer
d = 28.97 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: σy′ = −12.32 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Transformation equation for the y′ face
σy′ = 0 − (−10)·0.5 − (−20)·(−0.866) = −12.32 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = 12.32 MPa, σx′ + σy′ = 0 = σx + σy
Answer
σy′ = −12.32 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = 43.5 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Transformation equation
τθ = −−20·0.6428 + 40·0.766 = 43.5 MPa
Answer
τθ = 43.5 MPa
7.Axial stress & strain
Answer: P = 41.89 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
A₁ = 314.2 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·1257) = 1.592 × 10⁻⁶ mm/N
Load
P = 0.4/9.549 × 10⁻⁶ = 41888 N = 41.89 kN
Answer
P = 41.89 kN
8.Axial stress & strain
Answer: σ2 = 7.596 MPa
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Stiffnesses
k₁ = 200000·490.9/500 = 196350 N/mm
k₂ = 70000·2827/400 = 494801 N/mm
Share of segment 2
F2 = 30000·494801/691150 = 21477 N
Stress
σ2 = 21477/2827 = 7.596 MPa (compression)
Answer
σ2 = 7.596 MPa
9.Column buckling
Answer: Pcr = 3.052 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴
Effective length
KL = 1·3000 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·39761/3000² = 3052 N = 3.052 kN
Answer
Pcr = 3.052 kN
10.Column buckling
Answer: d = 79.34 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Required critical load
Pcr = n·P = 2·30000 = 60000 N
KL = 2·4000 = 8000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·60000·8000²/(π³·200000))^(1/4) = 79.34 mm
Answer
d = 79.34 mm
Check: the slenderness KL/r is 403.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.09999
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Logarithmic decrement
δ = (1/4)·ln(40/3.2) = 0.6314
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6314/√(39.48 + 0.3987) = 0.09999
Answer
ζ = 0.09999
12.Mechanical vibrations
Answer: fn = 2.531 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 12 + 5 = 17 kN/m
keq = 17·25/(17 + 25) = 10.12 kN/m
Natural frequency
ωn = √(10119/40) = 15.91 rad/s
fn = 15.91/(2π) = 2.531 Hz
Answer
fn = 2.531 Hz
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